What is the integral of (6x-11)/((x-1)^2) ?

The integral of (6x-11)/((x-1)^2) is 6ln|x-1|+5/(x-1)+C

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integral of (6x-11)/((x-1)^2)

\int \frac{6 x - 11}{( x - 1 ) ^{2}} d x

6 ln ∣ x - 1 ∣ + \frac{5}{x - 1} + C

Solution steps

\int \frac{6 x - 11}{( x - 1 ) ^{2}} d x

Expand

\frac{6 x - 11}{( x - 1 ) ^{2}} : \frac{6 x}{( x - 1 ) ^{2}} - \frac{11}{( x - 1 ) ^{2}}

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int \frac{6 x}{( x - 1 ) ^{2}} d x - \int \frac{11}{( x - 1 ) ^{2}} d x

\int \frac{6 x}{( x - 1 ) ^{2}} d x = 6 (ln ∣ x - 1 ∣ - \frac{1}{x - 1})

\int \frac{11}{( x - 1 ) ^{2}} d x = - \frac{11}{x - 1}

= 6 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - (- \frac{11}{x - 1})

Simplify

6 (ln ∣ x - 1 ∣ - \frac{1}{x - 1}) - (- \frac{11}{x - 1}) : 6 ln ∣ x - 1 ∣ + \frac{5}{x - 1}

= 6 ln ∣ x - 1 ∣ + \frac{5}{x - 1}

Add a constant to the solution

= 6 ln ∣ x - 1 ∣ + \frac{5}{x - 1} + C

l a pl a ce t r an s f or m x^{2}

\int \frac{1}{12 + 4 x - x ^{2}} d x

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d er i v a t i v e \frac{x ^{2} - 1}{( 2 x - 5 ) ^{2}}

\frac{d}{d x} (4 cos (sin (3 x)))

What is the integral of (6x-11)/((x-1)^2) ?
The integral of (6x-11)/((x-1)^2) is 6ln|x-1|+5/(x-1)+C